Evaluate the following integrals:

Here first of all convert secx in terms of $\cos x$

$\therefore$ We get

$\Rightarrow \sec x=\frac{1}{\cos x}, \sec 2 x=\frac{1}{\cos 2 x}$

$\therefore$ We get

$\Rightarrow \int \frac{\cos 2 x}{\cos x} d x$

We know

$\Rightarrow \int \frac{2 \cos ^{2} x-1}{\cos x} d x$

$\therefore$ We can write the above equation as

$\Rightarrow \int \frac{2 \cos ^{2} x-1}{\cos x} d x$

$\Rightarrow \int 2 \cos x d x-\int \frac{1}{\cos x} d x$

$\Rightarrow 2 \sin x-\int \sec x d x$

$\int \sec x d x=\ln |\sec x+\tan x|+c$

$\Rightarrow 2 \sin x-\ln |\sec x+\tan x|+c$